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2 years ago

14

A bike, a truck, and a train—all without passengers, motors, or engines—roll down the same hill. Put the vehicles in order from

the least amount of motion energy to the most.

1 answer:

Pepsi [2]2 years ago

4 0

Answer: WAIT WHATTTT i have that same test due today and the answer is in explanation

Explanation:

Bike, truck train. we are in the same school i think. Its imma say the incisal JMES Im Lusi i used to help in the library

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Jeff's body contains about 5.46 L of blood that has a density of 1060 kg/m3. Approximately 45.0% (by mass) of the blood is cells

barxatty [35]

Answer:

a

The mass of blood is $m= 5.7876kg$

b

The number of blood cells is $N_t=1.04*10^{13}$

Explanation:

From the question we are told that

The volume of blood is $V_b = 5.46 \ L = \frac{5.46}{1000} = 0.00546m^3$

The density of the blood is $\rho_b = 1060 kg/m^3$

% of blood that is cell is = 45.0%

% of the blood that is plasma is = 55.0%

density of blood cell is $\rho_d = 1125kg/m^3$

% of cell that are white is = 1%

% of cell that is red is = 99%

The diameter of the red blood cell is = $7.5 \mu m = 7.5*10^{-6}m$

The radius of the red blood cell is $= \frac{7.5*10^{-6}}{2} = 3.75*10^{-6}m$

Generally the mass is mathematically represented as

$m = \rho_b * V_b$

Substituting value

$m = 1060 * 0.00546$

$m= 5.7876kg$

Mass of cell is $m_c$ = 45% of m

= 0.45 * 5,7876

= 2.60442 kg

The volume of cells is $V_c$ = $\frac{m_c}{\rho_d}$

$= \frac{2.60442}{1125}$

$= 2.315 *10^{-3} m^3$

The volume of white blood cell is $V_w$ = 1% of volume of cells

$= \frac{1}{100} * 2.315*10^{-3}$

$= 2.315*10^{-5}m^3$

The volume of a single cell is $V_s = 4 \pi r^3$

$= 4*(3.142) * (3.75*10^{-6})^3$

$= 2.21*10^{-16}m^3$

The volume of red blood cells is $V_r = V_c - V_w$

$=2.315*10^{-3} - 2.315*10^{-5}$

$= 2.29*10^{-3}m^3$

The number of red blood cell is $= \frac{V_r}{V_s}$

$= \frac{2.29 *10^{-3}}{2.21*10^{-16}}$

$= 1.037*10^{13}$

The Number of white blood cell is $=\frac{V_w}{V_s}$

$= \frac{2.315 * 10^{-5}}{2.21*10^{-16}}$

$= 1.04*10^{11}$

The total number of blood cells is $N_t= 1.037*10^{13} + 1.04*10^{11}$

$N_t=1.04*10^{13}$

6 0

3 years ago

Read 2 more answers

If a car's speed triples, how does the momentum and kinetic energy of the

Harrizon [31]

Answer: When the car speed triples, momentum also triples but Kinetic energy increases 9 times or by 9 fold.

Explanation:

The momentum of a car (an object) is

p= mv

where

m is =the mass of the object( in this case car)

v is its= velocity

While the kinetic energy is is given by the formulae

K=1/2mv²

To determine how momentum and kinetic energy of the car changes when the speed of the object triples, We have that the new velocity,

v¹= 3v

So that the momentum change becomes

p¹=mv¹=m (3v)= 3mv

mv=p

therefore p¹= 3p

we can see that the momentum also triples.

And the kinetic energy change becomes

K¹=1/2m(v¹)²= 1/2m (3v)²

= 1/2m9v²= 1/2 x m x 9 x v²=9 x1/2mv²

1/2mv²=K

K¹= Kinetic energy = 9k

but Kinetic energy increases 9 times

7 0

2 years ago

Vector ~A has a negative x-component 3.07 units in length and a positive y-component 3.17 units in length. When a vector ~B = b1

luda_lava [24]

Answer:

a. 3.07 b. 1.26

Explanation:

Given that A = -3.07i + 3.17j and B = b1i + b1j and C = A + B = 0i + 4.43j

Since A + B = -3.07i + 3.17j + b1i + b2j

= (-3.07 + b1)i + (3.17 + b2)j

So,(-3.07 + b1)i + (3.17 + b2)j = 0i + 4.43j

Comparing components,

-3.07 + b1 = 0 (1) and 3.17 + b2 = 4.43 (2)

a. From (1), b1 = 3.07

b. From(2) b2 = 4.43 - 3.17 = 1.26

4 0

2 years ago

Can someone give me an example or explanation of calculating energy from voltage? Many thanks!

antoniya [11.8K]

Expression to calculate energy from voltage: E= V*Q where E= energy, V= voltage, and Q= charge

Additional help:
-To find the Voltage ( V )
[ V = I x R ] V (volts) = I (amps) x R (Ω)

-To find the Current ( I )
[ I = V ÷ R ] I (amps) = V (volts) ÷ R (Ω)

-To find the Resistance ( R )
[ R = V ÷ I ] R (Ω) = V (volts) ÷ I (amps)

I hope that helps to some extent-

7 0

2 years ago

When making maps of the large-scale universe, astronomers estimate distances to the vast majority of galaxies by using:

Vesnalui [34]

Answer:

<em>The comoving distance and the proper distance scale</em>

<em></em>

Explanation:

The comoving distance scale removes the effects of the expansion of the universe, which leaves us with a distance that does not change in time due to the expansion of space (since space is constantly expanding). The comoving distance and proper distance are defined to be equal at the present time; therefore, the ratio of proper distance to comoving distance now is 1. The scale factor is sometimes not equal to 1. The distance between masses in the universe may change due to other, local factors like the motion of a galaxy within a cluster. Finally, we note that the expansion of the Universe results in the proper distance changing, but the comoving distance is unchanged by an expanding universe.

4 0

2 years ago